Agenda

Sections 2.2, 2.3

Reminders

Read §2.4, 2.5, 2.8

Do problems for §2.2, 2.3

Homework 1 due Friday

Midterm Exam I on 1/23

Lab on Friday (Shapiro 2054)

Office hours Tues, Thurs3-4:30 pm (5852 East Hall)

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Definition: Separable DE

A DE of the form y ′ = f (x , y) is separable if f (x , y) can bewritten as the product of a function that only depends on xtimes another function that only depends on y . That is,

dy

dx= f (x , y) = p(x)q(y).

Example

Which DE’s are separable?

(a) sin (2x)dx + 3ydy = 0

(b) x2y ′ = y − xy

(c)√x2 − y 2 + y = xy ′

(d) y ′ = 1 + x + y 2 + xy 2

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Definition: Separable DE

A DE of the form y ′ = f (x , y) is separable if f (x , y) can bewritten as the product of a function that only depends on xtimes another function that only depends on y . That is,

dy

dx= f (x , y) = p(x)q(y).

Example

Which DE’s are separable?

(a) sin (2x)dx + 3ydy = 0

(b) x2y ′ = y − xy

(c)√x2 − y 2 + y = xy ′

(d) y ′ = 1 + x + y 2 + xy 2

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Suppose we have a separable DE of the form

dy

dx= p(x)q(y).

If we assume that q(y) is nonzero for the y values we’reinterested in, we can divide both sides by q(y).

1

q(y)

dy

dx= p(x)

Integrating both sides with respect to x gives∫1

q(y)

dy

dxdx =

∫p(x)dx =⇒

∫1

q(y)dy =

∫p(x)dx .

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

Show that the general solution to u′ = −k(u − T ) isu = T + ce−kt .

We begin with the DE

du

dt= −k(u − T ).

We separate the dependent and independent variables to get

du

u − T= −kdt.

(Note that the integration will be easier if the −k term stayson the right side.)

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

Show that the general solution to u′ = −k(u − T ) isu = T + ce−kt .

We begin with the DE

du

dt= −k(u − T ).

We separate the dependent and independent variables to get

du

u − T= −kdt.

(Note that the integration will be easier if the −k term stayson the right side.)

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Integrating both sides gives∫du

u − T= −

∫kdt.

Let w = u − T , so dw = du and the integral becomes∫dw

w= −

∫kdt.

Evaluating the integrals and subbing u − T in for w gives

ln |u − T | = −kt + c ,

where c is our constant of integration. Exponentiating bothsides gives

u − T = ±e−kt+c .

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Using our rules for exponents, we get

u = T ± ece−kt .

Since ±ec could be any arbitrary number (what about 0?), wecan say it’s just some arbitrary constant and replace it with c .That is,

u = T + ce−kt

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Definition: Explicit and Implicit Solutions

An explicit solution is a solution where the dependent variabley is isolated on one side. For example, y = −x/2. An implicitsolution is a solution where the dependent variable is notisolated. For example, sin (x + ey ) = 3y .

Example

y 2 − 2y = x3 + 2x2 + 2x + c is an implicit expression for thegeneral solution of the DE

dy

dx=

3x2 + 4x + 2

2(y − 1).

Is it possible to find an explicit expression for the solution thatsatisfies y(0) = −1? If so, what is it?

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Yes, y can be isolated because the general solution is aquadratic equation in terms y , and any quadratic equation canbe solved using the quadratic formula. To obtain an explicitexpression, let’s first find c by plugging in the initial condition.

(−1)2 − 2(−1) = 03 + 2 · 02 + 2 · 0 + c =⇒ c = 3

The implicit expression that satisfies the initial condition isthen

y 2 − 2y = x3 + 2x2 + 2x + 3.

To simplify the expression, let h(x) = x3 + 2x2 + 2x + 3, sowe have

y 2 − 2y − h(x) = 0.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

The quadratic equation gives us

y =2±√

4 + 4h(x)

2=⇒ y = 1±

√x3 + 2x2 + 2x + 4

To satisfy the initial condition, we must have

y = 1−√x3 + 2x2 + 2x + 4

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

Water leaks out of a cylindrical bucket at a rate proportionalto the square root of the water’s depth. If the water has aninitial depth of 1 meter and half of the water leaks out in 1minute, how long before the bucket is empty?

The DE being described is

dV

dt= −k

√h,

where V is the volume of water in the bucket and h is theheight of the water. Since V = Ah, where A is the area of thebottom of the bucket, we get that

dV

dt= A

dh

dt.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

Water leaks out of a cylindrical bucket at a rate proportionalto the square root of the water’s depth. If the water has aninitial depth of 1 meter and half of the water leaks out in 1minute, how long before the bucket is empty?

The DE being described is

dV

dt= −k

√h,

where V is the volume of water in the bucket and h is theheight of the water. Since V = Ah, where A is the area of thebottom of the bucket, we get that

dV

dt= A

dh

dt.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Plugging this into the DE gives

dh

dt= −B

√h,

where B = k/A. This is a separable DE, so we’ll separatevariables to get ∫

dh√h

= −∫

Bdt.

Integrating both sides gives

2√h = −Bt + c ,

where c is our constant of integration. To find c , we can plugin our initial condition.

2√

1 = −B · 0 + c =⇒ c = 2

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Since we know that h(1) = 1/2, we can find B .

2

√1

2= −B · 1 + 2 =⇒ B = 2−

√2

When the bucket is empty, h = 0, so we can now solve for thetime.

2√

0 = −(2−√

2)t + 2 =⇒ t =2

2−√

2≈ 3.41 minutes

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

A tank initially contains 10 lbs of salt dissolved in 100 gallonsof water. Fresh water is poured into the tank at a rate of 5gallons per minute. At the same time, water is drained fromthe tank at the same rate. Find the concentration of salt inthe tank as a function of time.

It’s usually a good idea to draw apicture that summarizes the problem.We will begin by constructing a DEfor the amount of salt in the tank.Since the amount of water in thetank remains constant, it should beeasy to then find the concentrationfrom the amount of salt.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

A tank initially contains 10 lbs of salt dissolved in 100 gallonsof water. Fresh water is poured into the tank at a rate of 5gallons per minute. At the same time, water is drained fromthe tank at the same rate. Find the concentration of salt inthe tank as a function of time.

It’s usually a good idea to draw apicture that summarizes the problem.We will begin by constructing a DEfor the amount of salt in the tank.Since the amount of water in thetank remains constant, it should beeasy to then find the concentrationfrom the amount of salt.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let Q(t) represent the amount of salt in the tank. We assumethat the salt is neither created nor destroyed, so the rate ofchange of salt in the tank, dQ/dt, is equal to the rate salt isflowing into the tank minus the rate salt is flowing out. Thatis,

dQ

dt= rate in− rate out.

Since only fresh water (no salt) is flowing into the tank, wehave

rate in = 0 lbs/min.

We can use dimensional analysis to find the rate out. Weknow that the concentration of salt in the tank is Q/100lbs/gal, so the amount of salt leaving the tank is

rate out =Q lbs

100 gal

5 gal

min=

Q

20lbs/min.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Therefore, the DE is

dQ

dt= − 1

20Q, Q(0) = 10

We can use separation of variables to solve the DE.∫dQ

Q= −

∫dt

20

ln |Q| = − 1

20t + c

Q = ±ece−t/20

Q = ce−t/20

Applying the initial conditions gives

Q = 10e−t/20.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Q = 10e−t/20

Recall that the problem originally asked for the concentrationof salt in the tank, not the total amount. Let C (t) representthe concentration of salt in the tank. Since the volume of thewater in the tank is constant, we have

C = Q/100 =1

10e−t/20 .

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

The population of mosquitoes in a certain area increases at arate proportional to the current population, and in the absenceof other factors, the population doubles each week. There are800,000 mosquitoes in the area initially, and predators (birds,bats, etc.) eat 30,000 mosquitoes/day. Determine thepopulation of mosquitoes in the area at any time.

Like the previous problem, we will need to determine the ratein and rate out for the mosquito population. Let M(t)represent the mosquito population at time t. According to theproblem, the population grows at a rate proportional to thecurrent population. So,

rate in = kM

where the constant k still needs to be determined.(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

The population of mosquitoes in a certain area increases at arate proportional to the current population, and in the absenceof other factors, the population doubles each week. There are800,000 mosquitoes in the area initially, and predators (birds,bats, etc.) eat 30,000 mosquitoes/day. Determine thepopulation of mosquitoes in the area at any time.

Like the previous problem, we will need to determine the ratein and rate out for the mosquito population. Let M(t)represent the mosquito population in thousands of mosquitoes.According to the problem, the population grows at a rateproportional to the current population. So,

rate in = kM thousand mosquitoes per day

where the constant k still needs to be determined.(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Since 30,000 mosquitoes are eaten per day,

rate out = 30 thousand mosquitoes per day

Therefore, our IVP is

dM

dt= kM − 30, M(0) = 800

Recall that the mosquito population doubles each week in theabsence of predators. We can use this information to find k .In the absence of predators, the IVP would then be

dM̃

dt= kM̃ , M̃(0) = M̃0,

where M̃0 is the initial number of mosquitoes.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

dM̃

dt= kM̃ , M̃(0) = M̃0

This DE can be solved using separation of variables.∫dM̃

M̃=

∫kdt

ln |M̃ | = kt + c

M̃ = ±ecekt

M̃ = cekt

Applying the initial condition gives

M̃ = M̃0ekt .

We know that after 1 week (7 days) the population doubles.Therefore,

2M̃0 = M̃0e7k =⇒ k = (ln 2)/7

.(Gary Marple) January 14th, 2019 Math 316: Differential Equations

k = (ln 2)/7

Now that we know k , we can plug it into the differentialequation for M to get

dM

dt=

ln 2

7M − 30, M(0) = 800.

This DE can be solved using separation of variables or anintegrating factor. We will use separation of variables since it’sprobably easier. We’ll start by factoring out the (ln 2)/7 term.

dM

dt=

ln 2

7

(M − 30

7

ln 2

)dM

dt=

ln 2

7

(M − 210

ln 2

)(Gary Marple) January 14th, 2019 Math 316: Differential Equations

dM

dt=

ln 2

7

(M − 210

ln 2

)Separating variables gives∫

dM

M − (210/ ln 2)=

∫ln 2

7dt

Notice that we only divided by the M − (210/ ln 2) term sinceit makes the integral on the left easier to evaluate. Evaluatingthe integrals gives

ln |M − (210/ ln 2)| =ln 2

7t + c

M − (210/ ln 2) = ±eceln 2

7t

M − (210/ ln 2) = ce ln (2t/7)

M = c2t/7 + (210/ ln 2)

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

M = c2t/7 + (210/ ln 2)

Recall that the initial condition was M(0) = 800. Pluggingthis in gives

800 = c20 + (210/ ln 2) =⇒ c = 800− 210

ln 2.

Therefore, our final answer is

M =

(800− 210

ln 2

)2t/7 +

210

ln 2

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

A body of mass m is projected away from the earth in adirection perpendicular to the earth’s surface with an initialvelocity v0. The gravitational force exerted by the earth isinversely proportional to the square of the distance betweenthe center of the earth and the object. Ignoring air resistance,what is the least initial velocity for which the body will notreturn to the earth? (This initial velocity is also known as theescape velocity.)

Again, sometimes it helps to create aquick drawing to visualize theproblem. Let R represent the radiusof the earth and x the distance fromthe surface of the earth to the object.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

A body of mass m is projected away from the earth in adirection perpendicular to the earth’s surface with an initialvelocity v0. The gravitational force exerted by the earth isinversely proportional to the square of the distance betweenthe center of the earth and the object. Ignoring air resistance,what is the least initial velocity for which the body will notreturn to the earth? (This initial velocity is also known as theescape velocity.)

Again, sometimes it helps to create aquick drawing to visualize theproblem. Let R represent the radiusof the earth and x the distance fromthe surface of the earth to the object.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Newton’s second law tells us that F = ma, or in this case

mdv

dt= Gravitational Force on the Object

The problem description said that the force due to gravity isinversely proportional to the square of the distance from thecenter of the earth to the object. Therefore,

mdv

dt= − k

(R + x)2

At this point, we should probably think through what exactlywe’re looking for. We could express dv/dt as d2x/dt2, but wehaven’t learned techniques for solving second order equationsyet. Instead, we’re going to use the chain rule to try andobtain a first order DE for v(x).

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

mdv

dt= − k

(R + x)2(∗)

The chain rule says that

dv

dt=

dv

dx

dx

dt,

but dx/dt = v . Therefore,

dv

dt=

dv

dxv .

Plugging this into (∗) gives us

mdv

dxv = − k

(R + x)2.

To find k , we can consider the gravitational force on thesurface of the earth. That is, we know that when x = 0, theforce due to gravity is approximately −mg , whereg = 9.8 m/s2 is the acceleration due to gravity. Therefore,

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

−mg = − k

(R + 0)2=⇒ k = mgR2

The IVP is then

mdv

dxv = − mgR2

(R + x)2, v(0) = v0,

or with cancellations

dv

dxv = − gR2

(R + x)2, v(0) = v0.

This can be solved using separation of variables.∫vdv = −

∫gR2

(R + x)2dx

v 2

2=

gR2

R + x+ c

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

v 2

2=

gR2

R + x+ c

Applying the initial conditions, v(0) = v0, gives

v 20

2=

gR2

R + 0+ c =⇒ c =

v 20

2− gR .

Our expression for v(x) is then

v 2

2=

gR2

R + x+

v 20

2− gR .

Suppose v0 is less than the escape velocity. Then, the objectwill reach its maximum height x = ξ when v = 0. Solving forv0 gives

02

2=

gR2

R + ξ+

v 20

2− gR =⇒ v0 =

√2gR

ξ

R + ξ

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

v0 =

√2gR

ξ

R + ξ(?)

Equation (?) gives us the initial velocity when the objectobtained a maximum distance of ξ from the surface of theearth. The escape velocity ve is found by letting ξ →∞.Therefore,

ve =√

2gR ≈ 11.2 km/s

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Example

Three tanks contain a mixture of salt and water. The firsttank initially has 30 lb of salt dissolved in 100 gallons of water.The second tank initially has 20 lb of salt dissolved in 200gallons of water, and the third tank initially has 10 lb of saltdissolved in 50 gallons of water. Suppose salt water flows intotank 1 at a rate of 3 gal/min with a concentration of 2 lbs ofsalt per gallon. Also, suppose that salt water flows from tank1 to tank 2, from tank 2 to tank 3, and out of tank 3 all at arate of 3 gal/min. Write a system of differential equationsthat models the amount of salt in each tank as a function oftime. Explain how you could go about solving such a system.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

Let x1,x2, and x3 represent theamount of salt in each tankrespectively.

x ′1 =2 lbs

gal

3 gal

min− x1 lbs

100 gal

3 gal

min

x ′2 =x1 lbs

100 gal

3 gal

min− x2 lbs

200 gal

3 gal

min

x ′3 =x2 lbs

200 gal

3 gal

min− x3 lbs

50 gal

3 gal

min

x1(0) = 30 lbs, x2(0) = 20 lbs, x3(0) = 10 lbs

x1 can be solved using separation of variables. Plug the solutionfor x1 into the DE for x2. x2 can be solved for using an integratingfactor. Then, plug the solution for x2 into the DE for x3. x3 canalso be solved for using an integrating factor.

(Gary Marple) January 14th, 2019 Math 316: Differential Equations

(Gary Marple) January 14th, 2019 Math 316: Differential Equations